Infimum of integral of open set

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I have already done part a and b. Part a is easy, for part b, i let the anti-derivative of f to be k(t)+c and arrive at the answer that F(f)= 1/2+ 2*k(1/2) - k(1). But i don't know how to do the next part, can anyone give me a hint? the question c ask me to show that the infimum of F is 0 and it is never attained on A.
 

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lahuxixi said:
I have already done part a and b. Part a is easy, for part b, i let the anti-derivative of f to be k(t)+c and arrive at the answer that F(f)= 1/2+ 2*k(1/2) - k(1). But i don't know how to do the next part, can anyone give me a hint? the question c ask me to show that the infimum of F is 0 and it is never attained on A.

Hint: For F(f) to be near zero then f needs to be near 0 on [0,1/2) and near 1 on (1/2,1].
 
lahuxixi said:
I have already done part a and b. Part a is easy, for part b, i let the anti-derivative of f to be k(t)+c and arrive at the answer that F(f)= 1/2+ 2*k(1/2) - k(1). But i don't know how to do the next part, can anyone give me a hint? the question c ask me to show that the infimum of F is 0 and it is never attained on A.

It's conceptually pretty easy. Both integrands are nonnegative. You can make it as small as you want by jumping from f=0 on most of [0,1/2] to f=1 on most of [1/2,1]. Is that enough of a hint?
 
but in that case, F(f) will be -1/2,right? I am sorry but i don't really understand the hint,can you give a more specific "hint"?(i know,im bad at this)
 
I don't know where you got that since "f" was not given explicitely. It sounds to me like you are integrating f from 0 to 1 rather than f from 0 to 1/2 and then 1- f from 1/2 to 1.

Suppose f(x)= 0 for 0\le x\le .4, f(x)= 5(x- .4) for .4\le x\le .6, f(x)= 1 for .6\le x\le 1. That is, f is 0 from 0 to .4, then the straight line from (.4, 0) to (.6, 1), then is 1 from .6 to 1. What is the integral of that?

Now, do the same thing for , say .45 instead of .4 and .55 instead of .6- f(x)= 0 for 0\le x\le .45, is the straight line from (.45, 0) to (.55, 1), then 1 from .55 to 1. What is the integral of that? (You don't even need to write out the formula for the function- the integral is the area of a right triangle.)

What happens as you keep moving those two points toward x= .5?
 
thank you very much for this, i have been trying to do this question for a whole day already.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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